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3 years ago

1. Calculate the molar mass of the substances involved in the reaction

1 answer:

Nookie1986 [14]3 years ago

3 0

The molar mass of a substance/chemical is the mass of the sample substance divided by the amount of substance in that sample.

Explanation:

In order to calculate the atomic mass of a substance we first obtain the atomic weight of the substance from the periodic table.
Then we count the number of atoms of the substance and multiply it with the individual atomic mass.

Molar mass of Citric Acid ( H3C6H5O7) is 192.1235 g/mol

It is calculated as :

Molar mass of C₆H₈O₇ = 6(atomic mass of C) + 8(atomic mass of H) + 7(atomic mass of O) = 6(12.0 g/mol) + 8(1.0 g/mol) + 7(16.0 g/mol) = 192.0 g/mol.

Molar mass of baking soda (NaHCO₃) is 84.0 g/mol

Molar mass of NaHCO₃ = (atomic mass of Na) + (atomic mass of H) + (atomic mass of C) + 3(atomic mass of O) = (23.0 g/mol) + (1.0 g/mol) + (12.0 g/mol) + 3(16.0 g/mol) = 84.0 g/mol.

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2 years ago

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2 years ago

A student placed 10.5 g of glucose (C6H12O6) in a volumetric fla. heggsk, added enough water to dissolve the glucose by swirling

aniked [119]

Answer: The mass of glucose in final solution is 0.420 grams

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$ .........(1)

Initial mass of glucose = 10.5 g

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$\text{Initial molarity of glucose}=\frac{10.5\times 1000}{180.16\times 100}\\\\\text{Initial molarity of glucose}=0.583M$

To calculate the molarity of the diluted solution, we use the equation:

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the molarity and volume of the concentrated glucose solution

$M_2\text{ and }V_2$ are the molarity and volume of diluted glucose solution

We are given:

$M_1=0.583M\\V_1=20.0mL\\M_2=?M\\V_2=0.5L=500mL$

Putting values in above equation, we get:

$0.583\times 20=M_2\times 500\\\\M_2=\frac{0.583\times 20}{500}=0.0233M$

Now, calculating the mass of final glucose solution by using equation 1:

Final molarity of glucose solution = 0.0233 M

Molar mass of glucose = 180.16 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

$0.0233=\frac{\text{Mass of glucose in final solution}\times 1000}{180.16\times 100}\\\\\text{Mass of glucose in final solution}=\frac{0.0233\times 180.16\times 100}{1000}=0.420g$

Hence, the mass of glucose in final solution is 0.420 grams

3 0

3 years ago

Determine the mass in grams of sucrose that must be added to 26 g of water to make a 1.15 m solution. Enter your answer to 2 dec

True [87]

Answer:

10.23 grams of sucrose should be added.

Explanation:

1.15 m means molality (moles of solute in 1kg of solvent)

1.15 moles of sucrose are contained in 1 kg of solvent (1000 g)

Let's determine the moles in our mass of solvent.

Firstly we convert the g to kg → 26 g . 1kg/1000g = 0.026 kg

m . mass (kg) = 1.15 mol/kg . 0.026kg → 0.0299 moles.

Finally we convert the moles to mass (mol . molar mass)

0.0299 mol . 342.3 g/mol = 10.23 g

4 0

3 years ago

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3 years ago